Respuesta :
Answer:
The answer is "0.9461"
Step-by-step explanation:
Given:
[tex]\int^1_0 \frac{sin(x)}{x} dx\\\\[/tex]
[tex]\int^1_0 \frac{sin(x)}{x} dx\\\\[/tex] ≈[tex]\sum^2_{n=0}\frac{(-1)^n x^{2n+1} (-1)^n (x-1)^n}{(2n + 1)!}[/tex]
[tex]\therefore[/tex]
[tex]\to \sin x = \sum^{\infty}_{n=0} (-1)^n\frac{x^{(2n+1)}}{(2n + 1)!}[/tex]
[tex]\because \\\\ \to \frac{\sin x}{x} =\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n+1-1}}{(2n+1)!}\\[/tex]
[tex]=\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n}}{(2n+1)!}[/tex]
The value is in between 0 and 1 then:
[tex]\to \int^1_0 \frac{sin(x)}{x} = =\sum^{2}_{n=0} \frac{(-1)^n}{ (2n+1) (2n+1)!}[/tex]
The above-given series is an alternative series, and it will give an error, when the nth term is bounded by its absolute value, that can be described as follows:
[tex]\to \frac{1}{(2n+3) (2n+3)!}< 0.0001\\\\\to (2n+3) (2n+3)!> 0.0001\\\\\to n\geq 2 \\[/tex]
So,
[tex]\int^1_0 \frac{sin(x)}{x} dx\\\\[/tex] ≈[tex]1 - \frac{1}{3 \cdot 3!}+\frac{1}{5 \cdot 5!}\\\\[/tex]
[tex]\approx 1 - \frac{1}{3 \cdot 6}+\frac{1}{5 \cdot 120}\\\\\approx 1 - \frac{1}{18}+\frac{1}{600}\\\\\approx 1 - \frac{1}{18}+\frac{1}{600}\\\\\approx \frac{18-1}{18}+\frac{1}{600}\\\\\approx \frac{17}{18}+\frac{1}{600}\\\\\approx \frac{1700+3}{1800}\\\\ \approx \frac{1703}{1800}\\\\\approx 0.9461[/tex]
The given series is an alternative series and will give an error when the nth term is bounded by its absolute value, then the approximation summation is equal to 0.9461.
What is a power series?
The finite series can be used as a polynomial with an infinite number of the term.
[tex]\int_{0}^{1} \dfrac{\sin x}{x} dx = \sum_{n=0}^{2} \dfrac{(-1)^{n}x^{2n+1}(-1)^{n}x^{n}}{(2n+1)!}[/tex]
Then we have
[tex]\Rightarrow \sin x = \sum_{n=0}^{\infty} (-1)^{n} \dfrac{x^{2n+1}}{(2n+1)!} \\\\\\\Rightarrow \dfrac{\sin x}{x} = \sum_{n=0}^{\infty} (-1)^{n} \dfrac{x^{2n+1-1}}{(2n+1)!}\\\\\\\Rightarrow \dfrac{\sin x}{x} = \sum_{n=0}^{\infty} (-1)^{n} \dfrac{x^{2n}}{(2n+1)!}[/tex]
Then the value is between 0 and 1, will be
[tex]\Rightarrow \int_0^1 \dfrac{\sin x}{x} = \sum_{n=0}^{2} \dfrac{(-1)^{n} }{(2n+1)(2n+1)!}[/tex]
The given series is an alternative series and will give an error when the nth term is bounded by its absolute value, then which can be described as
[tex]\Rightarrow \dfrac{1}{(2n+3)(2n+3)!} < 0.0001\\\\\\\Rightarrow (2n+3)(2n+3)! = > 0.0001\\\\\\\Rightarrow n > 2[/tex]
So,
[tex]\int _0^1 \dfrac{\sin x }{x} dx \approx 1- \dfrac{1}{3\times 3!} + \dfrac{1}{5\times 5!}\\\\\\\int _0^1 \dfrac{\sin x }{x} dx \approx 1- \dfrac{1}{3\times 6} + \dfrac{1}{5\times 120}\\\\\\\int _0^1 \dfrac{\sin x }{x} dx \approx 1- \dfrac{1}{18} + \dfrac{1}{600}\\\\\\\int _0^1 \dfrac{\sin x }{x} dx \approx 0.9461[/tex]
More about the power series link is given below.
https://brainly.com/question/14318966
